Question

f(x)在[0,2]连续,(0,2)内二阶可导,存在ξ∈(0,2),使f(0)-2f(1)+f(2)=f"(ξ)

Answer 1

f(x) = f(0) + f'(0)x + f''(ξ)x^2/2
f(1) = f(0) +f'(0) + f''悉或兆(ξ)/2 (1)
f(2)= f(0) +2f'(0) + 2f''团巧(ξ) (2)
(2)-2(1)
f(2)-2f(1) =-f(0)+f'睁租'(ξ)
f''(ξ)=f(0)-2f(1)+f(2)